All categories

4 years ago

A rectangular storage container with an open top is to have avolume

Mathematics

1 answer:

IRINA_888 [86]4 years ago

5 0

Answer:

Dimensions of the container:

x = 3 m

y = 6 m

h = 1.1 m

C(min) = 270 $

Step-by-step explanation:

Volume of storage container

V = 20 m³

Let "y" be the length and "x" the width then y = 2*x

V = x*y*h ⇒ V = 2*x²*h ⇒ 20 = 2*x²+h ⇒ h = 10/ x²

Costs:

Total cost = cost of base ( 5*2*x² ) + cost of side with base x ( 2*9*x*h) +

cost of side witn base y =2x (2*9*2x*h)

C(t) = 10*x² + 18*x*h + 36*x*h

C(x) = 10x² + 54*x*10/x² ⇒ C(x) 10*x² + 540 /x

Taking derivatives on both sides of the equation we get:

C´(x) = 20*x - 540/x²

C´(x) = 0 ⇒ 20*x - 540/x² = 0 ⇒ 2x - 54/x² = 0

2x³ - 54 = 0

x³ = 27 x = 3 m

Then y = 2*x ⇒ y = 2*3 y = 6 and h = 10 / x² h = 1.1 m

And the minimum cost is

C (min) = 10*x² + 540/x ⇒ C (min) = 90 + 180

C(min) = 270 $

You might be interested in

Simplify the expression 9m-7m+2m

tatuchka [14]

Simplify step-by-step.

9m−7m+2m

=9m+−7m+2m

Combine Like Terms:

=9m+−7m+2m

=(9m+−7m+2m)

=4m

4 0

3 years ago

Hi, how do we do this question?

Nutka1998 [239]

Answer:

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$

General Formulas and Concepts:

Algebra I

Terms/Coefficients
Factoring

Algebra II

Polynomial Long Division

Calculus

Differentiation

Derivatives
Derivative Notation

Derivative Property [Multiplied Constant]: $\displaystyle \frac{d}{dx} [cf(x)] = c \cdot f'(x)$

Derivative Property [Addition/Subtraction]: $\displaystyle \frac{d}{dx}[f(x) + g(x)] = \frac{d}{dx}[f(x)] + \frac{d}{dx}[g(x)]$

Basic Power Rule:

f(x) = cxⁿ
f’(x) = c·nxⁿ⁻¹

Integration

Integrals
Integration Constant C
Indefinite Integrals

Integration Rule [Reverse Power Rule]: $\displaystyle \int {x^n} \, dx = \frac{x^{n + 1}}{n + 1} + C$

Integration Property [Multiplied Constant]: $\displaystyle \int {cf(x)} \, dx = c \int {f(x)} \, dx$

Integration Property [Addition/Subtraction]: $\displaystyle \int {[f(x) \pm g(x)]} \, dx = \int {f(x)} \, dx \pm \int {g(x)} \, dx$

Logarithmic Integration

U-Substitution

Step-by-step explanation:

*Note:

You could use u-solve instead of rewriting the integrand to integrate this integral.

Step 1: Define

Identify

$\displaystyle \int {\frac{2x}{3x + 1}} \, dx$

Step 2: Integrate Pt. 1

[Integrand] Rewrite [Polynomial Long Division (See Attachment)]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\bigg( \frac{2}{3} - \frac{2}{3(3x + 1)} \bigg)} \, dx$
[Integral] Rewrite [Integration Property - Addition/Subtraction]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \int {\frac{2}{3}} \, dx - \int {\frac{2}{3(3x + 1)}} \, dx$
[Integrals] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}\int {} \, dx - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$
[1st Integral] Reverse Power Rule: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{3}\int {\frac{1}{3x + 1}} \, dx$

Step 3: Integrate Pt. 2

Identify variables for u-substitution.

Set u: $\displaystyle u = 3x + 1$
[u] Differentiate [Basic Power Rule]: $\displaystyle du = 3 \ dx$

Step 4: Integrate Pt. 3

[Integral] Rewrite [Integration Property - Multiplied Constant]: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{3}{3x + 1}} \, dx$
[Integral] U-Substitution: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}\int {\frac{1}{u}} \, du$
[Integral] Logarithmic Integration: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|u| + C$
Back-Substitute: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{2}{3}x - \frac{2}{9}ln|3x + 1| + C$
Factor: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = -2 \bigg( \frac{1}{9}ln|3x + 1| - \frac{x}{3} \bigg) + C$
Rewrite: $\displaystyle \int {\frac{2x}{3x + 1}} \, dx = \frac{-2(ln|3x + 1| - 3x)}{9} + C$